Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main(){
  int n;
  cin>>n;
  long double p[n+1];
  long double dp[n+1][n+1];
  dp[0][0]=1;
  for(int i=1;i<=n;i++){
    cin>>p[i];
    dp[i][0]=0;
    for(int j=1;j<=i;j++){
      dp[i][j-1]=dp[i][j-1]+(1-p[i])*dp[i-1][j-1];
      dp[i][j]=p[i]*dp[i-1][j-1];
    }
  }
  long double ans=0;
  for(int i=n/2+1;i<=n;i++)
    ans+=dp[n][i];
  cout<<fixed<<setprecision(9)<<ans<<'\n';
}